Datasets:

---

math-eval

/

TAL-SCQ5K

like 57

[ "2011年六年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$0.\\dot{5}\\dot{1}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{5}{9}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{24}{47}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{13}{25}$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->比较与估算->比较大小综合" ]

[ "六个数的大小关系为$$\\frac{24}{47} \\textless{} 0.5\\dot{1} \\textless{} 0.\\dot{5}\\dot{1} \\textless{} \\frac{13}{25} \\textless{} \\frac{5}{9} \\textless{} \\frac{2}{3}$$，根据已知条件，$$0.5\\dot{1}$$为从小到大第四个，所以从大到小排列第四个数为$$0.\\dot{5}\\dot{1}$$。 " ]

[ "六年级上学期其它第12讲", "2011年北京五年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$9$$ " } ], [ { "aoVal": "D", "content": "$$12$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "$$4\\times 3=12$$（千米），通过画图，我们发现甲走了一个全程多了回来那一段，就是距$$B$$地的$$3$$千米，所以全程是$$12-3=9$$（千米），两次相遇点相距$$9-\\left( 3+4 \\right)=2$$（千米）． " ]

[ "2006年第4届创新杯五年级竞赛初赛B卷第10题", "2006年五年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "3 " } ], [ { "aoVal": "B", "content": "4 " } ], [ { "aoVal": "C", "content": "5 " } ], [ { "aoVal": "D", "content": "6 " } ] ]

[ "拓展思维->拓展思维->组合模块->抽屉原理->最不利原则" ]

[ "每一排坐的人数都不同最多可坐$$\\left( 7+30 \\right)\\div 2\\times 24=444$$人 每一排坐的人数有$$2$$个相同最多可坐$$30\\times 2+29\\times 2+......+19\\times 2=588 \\textless{} 650$$ 每一排坐的人数有$$3$$个相同最多可坐$$30\\times 3+29\\times 3+......+23\\times 3=636 \\textless{} 650$$ 每一排坐的人数有$$4$$个相同最多可坐$$30\\times 4+29\\times 4+......+25\\times 4=660\\textgreater650$$ 故，至少有$$4$$排座位上人数相同 " ]

[ "2006年第4届创新杯五年级竞赛初赛B卷第6题", "2006年五年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "1000元 " } ], [ { "aoVal": "B", "content": "900元 " } ], [ { "aoVal": "C", "content": "990元 " } ], [ { "aoVal": "D", "content": "980元 " } ] ]

[ "拓展思维->拓展思维->应用题模块->平均数问题->综合题->综合题普通型" ]

[ "将第一种方案购买件数同时都扩大2倍，则购买货物$${{A}_{1}}$$、$${{A}_{2}}$$、$${{A}_{3}}$$、$$\\cdots$$、$${{A}_{10}}$$的件数依次为$$2$$，$$6$$，$$8$$，$$10$$，$$12$$，$$14$$，$$16$$，$$18$$，$$20$$，$$22$$件，需人民币$$1995\\times 2=3990$$(元)，所以各购买一件共需人民币$$3990-3000=990$$(元). " ]

[ "2019年第24届YMO一年级竞赛决赛第6题3分" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

[ "拓展思维->拓展思维->计数模块->枚举法综合->整数分拆->简单拆分->加法拆数（指定个数）", "Overseas Competition->知识点->计数模块->枚举法综合->枚举法" ]

[ "两位数的十位数字与个位数字之和为$$12$$， 而数字均为$$0\\sim 9$$，故枚举得： $$93$$，$$84$$，$$75$$，$$66$$，$$57$$，$$48$$，$$39$$共$$7$$个． 故选：$$\\text{C}$$． " ]

[ "2017年全国亚太杯五年级竞赛初赛第6题" ]

[ [ { "aoVal": "A", "content": "$$16$$ " } ], [ { "aoVal": "B", "content": "$$18$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$24$$ " } ] ]

[ "知识标签->拓展思维->应用题模块->比例应用题->按比分配" ]

[ "设小刚的藏书数量为$$8$$份，则小明和小红的藏书数量分别为$$3$$份和$$5$$份所以一份为$$48\\div \\left( 3+5+8 \\right)=3$$（本）． 所以小刚的藏书数量是$$8\\times 3=24$$ （本）． " ]

[ "2017年全国华杯赛竞赛初赛模拟试卷2第4题" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$10$$ " } ], [ { "aoVal": "C", "content": "$$15$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ] ]

[ "拓展思维->思想->枚举思想" ]

[ "设整数部分为$$5$$的小数为$$a$$，则$$5\\leqslant a\\textless6$$，整数部分为$$9$$的小数为$$b$$，则$$9\\leqslant b\\textless10$$，那么$$5\\times9\\leqslant a\\times b\\textless6\\times10$$，所以$$a\\times b$$的整数部分可以等于$$45$$至$$59$$之间的任意整数，共$$15$$种可能． " ]

[ "2011年第7届全国新希望杯六年级竞赛第3题4分" ]

[ [ { "aoVal": "A", "content": "$$3:27$$ " } ], [ { "aoVal": "B", "content": "$$4:17$$ " } ], [ { "aoVal": "C", "content": "$$5:14$$ " } ], [ { "aoVal": "D", "content": "$$6:22$$ " } ] ]

[ "知识标签->拓展思维->行程模块->时钟问题->时钟上的追及问题->已知时间求角度" ]

[ "b选项：$4\\times30-（6-0.5）=26.5^{\\circ}$，离$$30$$度最近，故选$$b$$． " ]

[ "2011年第9届全国创新杯小学高年级六年级竞赛第9题" ]

[ [ { "aoVal": "A", "content": "$$3$$个 " } ], [ { "aoVal": "B", "content": "$$4$$个 " } ], [ { "aoVal": "C", "content": "$$5$$个 " } ], [ { "aoVal": "D", "content": "$$2$$个 " } ] ]

[ "知识标签->拓展思维->数论模块->高斯记号->高斯记号的复杂应用" ]

[ "根据高斯函数的性质，可得$4\\leqslant\\dfrac{3x+7}{7}\\lt5$，解得$$7\\leqslant x\\leqslant 9$$，对应的整数共$$3$$个数． " ]

[ "2015年第11届全国新希望杯五年级竞赛复赛第1题" ]

[ [ { "aoVal": "A", "content": "$$106$$分 " } ], [ { "aoVal": "B", "content": "$$110$$分 " } ], [ { "aoVal": "C", "content": "$$114$$分 " } ], [ { "aoVal": "D", "content": "$$116$$分 " } ] ]

[ "拓展思维->拓展思维->应用题模块->平均数问题->综合题->综合题普通型" ]

[ "$$105.4\\times 5-120-90-102-99=116$$． " ]

[ "2017年全国华杯赛小学高年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{p}$$代表一个特定的数 " } ], [ { "aoVal": "B", "content": "$${{p}^{2}}$$代表一个特定的数 " } ], [ { "aoVal": "C", "content": "$$p$$代表一个尚未确定的数 " } ], [ { "aoVal": "D", "content": "$$\\frac{1-p}{p}$$代表一个特定的数 " } ] ]

[ "拓展思维->拓展思维->计算模块->小数->循环小数->循环小数的概念" ]

[ "如果圆周率$$\\pi $$，黄金分割数量个特定的无限不循环小数，则$$\\frac{1}{p}$$，$${{p}^{2}}$$和$$\\frac{1-p}{p}$$都是特定的数． " ]

[ "2014年全国迎春杯六年级竞赛复赛第2题" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

[ "拓展思维->思想->整体思想" ]

[ "由新定义：$$n!=1\\times 2\\times 3\\times \\ldots\\times n$$得： $$2014!=1\\times 2\\times 3\\times 4\\times 5\\times \\ldots\\times 2021\\times 2022$$ $$=1\\times 3\\times 4\\times 6\\times 7\\times 8\\times \\ldots\\times 2021\\times 2022\\times 10$$ 所以$$1\\times 3\\times 4\\times 6\\times 7\\times 8\\times \\ldots\\times 2021\\times 2022\\times 10$$是$$10$$的倍数， 所以$$2022!$$的个位数为$$0$$； $$3!=1\\times 2\\times 3=6$$ 所以$$2022!-3!$$的个位数也就为：$$10-6=4$$． " ]

[ "2018年第22届广东世界少年奥林匹克数学竞赛三年级竞赛决赛第4题5分" ]

[ [ { "aoVal": "A", "content": "趣 " } ], [ { "aoVal": "B", "content": "味 " } ], [ { "aoVal": "C", "content": "题 " } ] ]

[ "拓展思维->拓展思维->应用题模块->周期问题->基本排列的周期问题" ]

[ "观察题干可知，$$5$$个字一个循环周期，分别按照数、学、趣、味、题，依次排列，据此求出第$$2019$$个字是第几个循环周期的第几个字即可解答问题． $$2019\\div 5=403$$（组）$$\\cdots \\cdots 4$$（个）， 所以第$$2019$$个字是第$$403$$循环周期的第$$4$$个字，是``味''． 答：第$$2019$$个字是味． 故选：$$\\text{B}$$． " ]

[ "2009年五年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "1995 " } ], [ { "aoVal": "B", "content": "1990 " } ], [ { "aoVal": "C", "content": "1992 " } ], [ { "aoVal": "D", "content": "1989 " } ] ]

[ "拓展思维->拓展思维->应用题模块->年龄问题->年龄问题之当当型->年龄轴" ]

[ "设父亲年龄为x岁，兄弟二人的年龄分别为a岁、b岁，$$\\left( a\\textgreater b \\right)$$.则$$x=2\\left( a+b \\right)=7\\left( a-b \\right)$$，且$$x+a+b=84$$.将$$a+b=84-x$$代入$$x=2\\left( a+b \\right)$$得$$x=2\\left( 84-x \\right)$$，$$3x=84\\times 2$$.从而$$x=56$$，所以$$a+b=28$$，$$ab=8$$.因此$$a=18$$，$$b=10$$，哥哥生于1990年. " ]

[ "2021年新希望杯六年级竞赛初赛第11题5分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{99}{100}\\textgreater{} \\frac{100}{101}\\textgreater{} \\frac{199}{201}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{199}{201}\\textgreater{} \\frac{100}{101}\\textgreater{} \\frac{99}{100}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{99}{100}\\textgreater{} \\frac{199}{201}\\textgreater{} \\frac{100}{101}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{100}{101}\\textgreater{} \\frac{199}{201}\\textgreater\\frac{99}{100}$$ " } ], [ { "aoVal": "E", "content": "$$\\frac{100}{101}\\textgreater{} \\frac{99}{100}\\textgreater{} \\frac{199}{201}$$ " } ] ]

[ "拓展思维->能力->数感认知->分数数字加工" ]

[ "$$\\frac{99}{100}=1- \\frac{1}{100}=1- \\frac{2}{200}$$， $$\\frac{100}{101}=1- \\frac{1}{101}=1- \\frac{2}{202}$$， $$\\frac{199}{201}=1- \\frac{2}{201}$$， 因为$$\\frac{2}{202}\\textless{} \\frac{2}{201}\\textless{} \\frac{2}{200}$$， 所以$$1- \\frac{2}{202}\\textgreater1- \\frac{2}{201}\\textgreater1- \\frac{2}{200}$$， 即$$\\frac{100}{101}\\textgreater{} \\frac{199}{201}\\textgreater{} \\frac{99}{100}$$． 故选$$\\text{D}$$． " ]

[ "2020年福建河仁杯六年级竞赛初赛A卷第1题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{2018}{2019}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{2020}{2021}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{2019}{2020}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{2021}{2022}$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\frac{3}{{{1}^{2}}+1}-\\frac{5}{{{2}^{2}}+2}+\\frac{7}{{{3}^{2}}+3}-\\frac{9}{{{4}^{2}}+4}+\\cdots +\\frac{4039}{{{2019}^{2}}+2019}-\\frac{4041}{{{2020}^{2}}+2021}$$ $$=\\frac{3}{1\\times 2}-\\frac{5}{2\\times 3}+\\frac{7}{3\\times 4}-\\frac{9}{4\\times 5}+\\cdots +\\frac{4039}{2019\\times 2020}-\\frac{4041}{2020\\times 2021}$$ $$=\\left( 1+\\frac{1}{2} \\right)-\\left( \\frac{1}{2}+\\frac{1}{3} \\right)+\\left( \\frac{1}{3}+\\frac{1}{4} \\right)-\\left( \\frac{1}{4}+\\frac{1}{5} \\right)+\\cdots +\\left( \\frac{1}{2019}+\\frac{1}{2020} \\right)-\\left( \\frac{1}{2020}+\\frac{1}{2021} \\right)$$ $$=1+\\frac{1}{2}-\\frac{1}{2}-\\frac{1}{3}+\\frac{1}{3}+\\frac{1}{4}-\\frac{1}{4}-\\frac{1}{5}+\\cdots +\\frac{1}{2019}+\\frac{1}{2020}-\\frac{1}{2020}-\\frac{1}{2021}$$ $$=1-\\frac{1}{2021}$$ $$=\\frac{2020}{2021}$$． 故选$$\\text{B}$$． " ]

[ "2014年华杯赛六年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$14$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ], [ { "aoVal": "D", "content": "$$17$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->时间问题->时间计算" ]

[ "解：依题意可知： 上午十点对好表，标准钟每小时走$$60$$格，小华的表快$$4$$分钟每小时走$$64$$格。路程比例为$$15:16$$。 当小华的表为下午$$2$$点时，小华的表走了$$4$$圈共$$240$$格。 根据比例关系设标准钟走的路程为$$x$$，则有$$15:16=x:240$$，解得$$x=225$$。 $$240-225=15$$（分钟）。 故选：B " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛三年级竞赛决赛第1题5分" ]

[ [ { "aoVal": "A", "content": "一 " } ], [ { "aoVal": "B", "content": "二 " } ], [ { "aoVal": "C", "content": "三 " } ], [ { "aoVal": "D", "content": "四 " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "从$$2019$$年的$$1$$月$$1$$日至$$2019$$年的$$10$$月$$1$$日共：$$31+28+31+30+31+30+31+31+30+1=274$$（天）， 根据``二、三、四、五、六、日、一''的周期规律：$$274\\div 7=39$$（周）$$\\cdots \\cdots 1$$（天）． 所以$$10$$月$$1$$日是这个周期的第$$1$$天，即星期二． 故选$$\\text{B}$$． " ]

[ "2018年第22届广东世界少年奥林匹克数学竞赛五年级竞赛决赛第1题5分" ]

[ [ { "aoVal": "A", "content": "$$96$$ " } ], [ { "aoVal": "B", "content": "$$97$$ " } ], [ { "aoVal": "C", "content": "$$98$$ " } ] ]

[ "拓展思维->能力->数据处理" ]

[ "四位同学的总分为：$$95\\times4=380$$（分）， 去掉宇宇总分：$$94\\times3=282$$（分）， 故，宇宇的分数为：$$380-282=98$$（分）． 故选：$$\\text{C}$$． " ]

[ "2019年第23届广东世界少年奥林匹克数学竞赛二年级竞赛初赛第6题5分" ]

[ [ { "aoVal": "A", "content": "$$32$$ " } ], [ { "aoVal": "B", "content": "$$36$$ " } ], [ { "aoVal": "C", "content": "$$40$$ " } ], [ { "aoVal": "D", "content": "$$42$$ " } ] ]

[ "拓展思维->能力->数据处理" ]

[ "根据题目叙述倒推，第二天又摘了余下桃子的一半多$$1$$个，这时树上还有$$9$$个桃子，所以第一天摘完还剩桃子$$\\left(9+1\\right)\\div\\frac{1}{2}=20$$（个），因为第一天摘了树上桃子的一半，所以原来树上有$$20\\div\\frac{1}{2}=40$$（个），所以答案为$$40$$个． 故选：$$\\text{C}$$． " ]

[ "2017年河南郑州联合杯小学高年级六年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$20 \\% $$ " } ], [ { "aoVal": "B", "content": "$$25 \\% $$ " } ], [ { "aoVal": "C", "content": "$$30 \\% $$ " } ], [ { "aoVal": "D", "content": "$$22.5 \\% $$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->分百应用题->量率对应已知单位1" ]

[ "甲数比乙数少$$\\frac{1}{5}$$，设乙为5份，甲为4份，乙数比甲数多$$\\frac{5-4}{4}=\\frac{1}{4}=25 \\%$$． " ]

[ "2016年IMAS小学高年级竞赛第二轮检测试题第1题4分" ]

[ [ { "aoVal": "A", "content": "$$2433$$ " } ], [ { "aoVal": "B", "content": "$$2970$$ " } ], [ { "aoVal": "C", "content": "$$2973$$ " } ], [ { "aoVal": "D", "content": "$$3030$$ " } ], [ { "aoVal": "E", "content": "$$3033$$ " } ] ]

[ "拓展思维->能力->运算求解->程序性计算" ]

[ "$$666+669+699+999$$ $$=670+670+700+1000-4-1-1-1$$ $$=3040-7$$ $$=3033$$． " ]

[ "2019年第24届YMO一年级竞赛决赛第6题3分" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ] ]

[ "Overseas Competition->知识点->计数模块->枚举法综合->枚举法", "拓展思维->能力->逻辑分析" ]

[ "两位数的十位数字与个位数字之和为$$12$$， 而数字均为$$0\\sim 9$$，故枚举得： $$93$$，$$84$$，$$75$$，$$66$$，$$57$$，$$48$$，$$39$$共$$7$$个． 故选：$$\\text{C}$$． " ]

[ "其它改编自2014年全国希望杯六年级竞赛初赛第6题" ]

[ [ { "aoVal": "A", "content": "$$\\frac{20}{99}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{30}{99}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{40}{99}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{50}{99}$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "假设分数的分母为$$y$$，分子分别为$$2x$$，$$3x$$，$$4x$$，根据题意$$\\frac{2x}{y}+\\frac{3x}{y}+\\frac{4x}{y}=\\frac{10}{11}$$，所以解出最大分数为$$\\frac{4x}{y}=\\frac{40}{99}$$． " ]

[ "1989年华杯赛六年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$14$$ " } ], [ { "aoVal": "B", "content": "$$15$$ " } ], [ { "aoVal": "C", "content": "$$16$$ " } ] ]

[ "拓展思维->拓展思维->组合模块->抽屉原理->最不利原则" ]

[ "如果在最不利的情况下能完成目标，则保证在任何情况下都能完成。最不利的情况是：抽的前$$12$$张是$$4$$种花色各$$3$$张，再抽两张王牌，这时抽第$$15$$张，无论是哪种花色，都能保证凑成$$4$$张牌同一花色。所以至少要抽$$15$$张牌，才能保证有四张牌是同一花色的。 " ]

[ "2020年希望杯五年级竞赛模拟第14题", "2020年新希望杯五年级竞赛第14题" ]

[ [ { "aoVal": "A", "content": "$$3200$$ " } ], [ { "aoVal": "B", "content": "$$3600$$ " } ], [ { "aoVal": "C", "content": "$$32000$$ " } ], [ { "aoVal": "D", "content": "$$36000$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$$0.004186\\times 8812345.321\\approx 0.004\\times 9000000=36000$$． 故选$$\\text{D}$$． " ]

[ "2015年湖北武汉世奥赛五年级竞赛模拟训练题（三）第3题" ]

[ [ { "aoVal": "A", "content": "$$51$$ " } ], [ { "aoVal": "B", "content": "$$67$$ " } ], [ { "aoVal": "C", "content": "$$86$$ " } ], [ { "aoVal": "D", "content": "$$100$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "因为$$1$$，$$4$$，$$7$$都是除以$$3$$余$$1$$的数，小敏答对$$20$$道题，$$\\left( 20\\times 1 \\right)\\div 3=6\\ldots \\ldots 2$$，所以她的分数也一定是除以$$3$$余$$2$$的数，综合比较$$A$$，$$B$$，$$C$$，$$D$$四个选项，不难发现，只有$$C$$选项符合题意，即小敏可能的得分是$$86$$． " ]

[ "2018年迎春杯小学高年级竞赛决赛A卷第9题12分" ]

[ [ { "aoVal": "A", "content": "$$60$$ " } ], [ { "aoVal": "B", "content": "$$63$$ " } ], [ { "aoVal": "C", "content": "$$72$$ " } ], [ { "aoVal": "D", "content": "$$81$$ " } ], [ { "aoVal": "E", "content": "$$96$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想", "海外竞赛体系->知识点->数论模块->质数与合数->特殊质数运用->特殊质数2" ]

[ "设$$A=151514141313\\cdots 0707$$， （$$1$$）由四位数截断法知$$A$$是$$9999$$的倍数，不妨设$$A=9999B$$，则$$B$$是个$$32$$位数； （$$2$$）原式$$=\\underbrace{20182018\\cdots 2018}_{18个2018}\\times 9999\\times B=\\underbrace{99999999\\cdots 9999}_{18个9999}\\times 2018\\times B=\\underbrace{99\\cdots 9}_{72个9}\\times 2018\\times B$$ 因为$$2018\\times B$$是一个不超过$$72$$位的数，无论它是多少，$$\\underbrace{99\\cdots 9}_{72个9}\\times 2018\\times B$$的计算结果中都是$$72$$个奇数数字． " ]

[ "2020年广东广州羊排赛六年级竞赛第7题3分" ]

[ [ { "aoVal": "A", "content": "$$\\frac{5}{7}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{2}{13}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{3}{4}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{10}{17}$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "通分子，都变成$$30$$去比较大小 则从大到小排列排在第二位的是$$\\frac{5}{7}$$． 故选$$\\text{A}$$． " ]

[ "2014年全国迎春杯五年级竞赛初赛第3题" ]

[ [ { "aoVal": "A", "content": "$$18$$ " } ], [ { "aoVal": "B", "content": "$$19$$ " } ], [ { "aoVal": "C", "content": "$$20$$ " } ], [ { "aoVal": "D", "content": "$$21$$ " } ] ]

[ "拓展思维->思想->逐步调整思想" ]

[ "$$48\\div 2.5=19\\cdots 0.5$$，$$19+1=20$$(辆)． " ]

[ "2013年第11届全国创新杯五年级竞赛第6题5分" ]

[ [ { "aoVal": "A", "content": "$$1$$与$$101$$之间的$$2$$倍数 " } ], [ { "aoVal": "B", "content": "$$1$$与$$101$$之间的$$3$$倍数 " } ], [ { "aoVal": "C", "content": "$$1$$与$$101$$之间的$$4$$倍数 " } ], [ { "aoVal": "D", "content": "$$1$$与$$101$$之间的$$6$$倍数 " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "根据等差数列相关概念可知，一个等差数列的平均数，其实就是这个数列的首项与末项的平均数，简化计算后易知$$1$$与$$101$$之间$$4$$的倍数的平均数最大，其值为$$(4+100)\\div 2=52$$． " ]

[ "2014年广东广州羊排赛六年级竞赛第9题1分" ]

[ [ { "aoVal": "A", "content": "$$20 \\%$$ " } ], [ { "aoVal": "B", "content": "$$25 \\%$$ " } ], [ { "aoVal": "C", "content": "$$27.5 \\%$$ " } ], [ { "aoVal": "D", "content": "$$30 \\%$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->浓度问题->浓度基本题型->已知溶质溶液求浓度" ]

[ "最后浓度变为$$\\frac{100\\times 20 \\%+10}{100+10+10}\\times 100 \\%=25 \\%$$． " ]

[ "2015年第4届广东广州羊排赛六年级竞赛第10题1分" ]

[ [ { "aoVal": "A", "content": "左，$$2$$ " } ], [ { "aoVal": "B", "content": "左，$$3$$ " } ], [ { "aoVal": "C", "content": "右，$$2$$ " } ], [ { "aoVal": "D", "content": "右，$$3$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->加减法应用->加减法应用顺口溜" ]

[ "相当于共向右走了$$3+4+3=10$$（步），共向左走了$$5+2=7$$（步），$$10-7=3$$（步），相当于在起点右方$$3$$步的地方． 故选$$\\text{D}$$． " ]

[ "2017年全国华杯赛小学中年级竞赛初赛模拟第1题", "小学中年级三年级上学期其它" ]

[ [ { "aoVal": "A", "content": "$$150$$ " } ], [ { "aoVal": "B", "content": "$$140$$ " } ], [ { "aoVal": "C", "content": "$$130$$ " } ], [ { "aoVal": "D", "content": "$$120$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->和差倍问题->和差问题->二量和差问题->明和暗差" ]

[ "因为甲给乙$$40$$元，甲和乙钱数相等，所以甲$$200+40\\times$$ 2=280（元），$$280\\div$$ 2=140（元） " ]

[ "2020年新希望杯六年级竞赛（2月）第25题" ]

[ [ { "aoVal": "A", "content": "$$\\textgreater$$ " } ], [ { "aoVal": "B", "content": "$$\\textless{}$$ " } ], [ { "aoVal": "C", "content": "$$=$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "$$\\frac{1}{{{2}^{2}}}+\\frac{1}{{{3}^{2}}}+\\cdots +\\frac{1}{{{2020}^{2}}} ~\\textless{} ~\\frac{1}{1\\times 2}+\\frac{1}{2\\times 3}+\\cdots +\\frac{1}{1999\\times 2020}$$ $$=1-\\frac{1}{2020}=\\frac{2019}{2020}$$． ∴$$1+\\frac{1}{{{2}^{2}}}+\\frac{1}{{{3}^{2}}}+\\cdots +\\frac{1}{{{2020}^{2}}} ~\\textless{} ~1+\\frac{2019}{2020} ~\\textless{} ~2$$． 故选$$\\text{B}$$． " ]

[ "2017年第15届湖北武汉创新杯六年级竞赛初赛第7题" ]

[ [ { "aoVal": "A", "content": "$$4$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ] ]

[ "拓展思维->能力->抽象概括" ]

[ "考虑最不利情况，先摸到一双，两只，再每种颜色各一只，共三只，再任意摸一只就可以配成两双，所以一共至多$$6$$只． " ]

[ "2016年全国小学生数学学习能力测评四年级竞赛复赛第7题3分" ]

[ [ { "aoVal": "A", "content": "$$60006000$$ " } ], [ { "aoVal": "B", "content": "$$60606060$$ " } ], [ { "aoVal": "C", "content": "$$6006060$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->位值原理与进制->数与数字->数、数位、数字的认识" ]

[ "$$60006000$$读作六千万六千，没有读出``零''； $$60606060$$读作六千零六十万六千零六十，读了两个``零''； $$6006060$$读作六百万六千零六十，读了一个``零''． 故选：$$\\text{B}$$． " ]

[ "2018年IMAS小学中年级竞赛（第一轮）第1题3分" ]

[ [ { "aoVal": "A", "content": "$$1900$$ " } ], [ { "aoVal": "B", "content": "$$1919$$ " } ], [ { "aoVal": "C", "content": "$$2900$$ " } ], [ { "aoVal": "D", "content": "$$2919$$ " } ], [ { "aoVal": "E", "content": "$$3800$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->整数->整数提取公因数->整数乘法巧算之提取公因数（普通型）" ]

[ "$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde19\\times 1+19\\times 3+19\\times 5+19\\times 7+\\cdots +19\\times 19$$ $$=19\\times \\left( 1+3+5+7+\\cdots +19 \\right)$$ $$=19\\times \\frac{\\left( 1+19 \\right)\\times 10}{2}$$ $$=19\\times 100$$ $$=1900$$． 故选$$\\text{A}$$． " ]

[ "2018年第6届湖北长江杯六年级竞赛初赛A卷第6题3分" ]

[ [ { "aoVal": "A", "content": "成正比例 " } ], [ { "aoVal": "B", "content": "成反比例 " } ], [ { "aoVal": "C", "content": "不成比例 " } ], [ { "aoVal": "D", "content": "无选项 " } ] ]

[ "拓展思维->拓展思维->几何模块->立体图形->圆柱与圆锥->圆锥的基本公式" ]

[ "圆锥的底面周长表示为$$2\\pi r$$， 圆锥的体积表示为$$\\frac{1}{3}\\pi {{r}^{2}}$$， 两者之间没有任何关系，所以故选$$\\text{C}$$． " ]

[ "2011年第7届全国新希望杯小学高年级六年级竞赛第6题4分" ]

[ [ { "aoVal": "A", "content": "$$108$$ " } ], [ { "aoVal": "B", "content": "$$96$$ " } ], [ { "aoVal": "C", "content": "$$88$$ " } ], [ { "aoVal": "D", "content": "$$81$$ " } ], [ { "aoVal": "E", "content": "$$54$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "先算卡片的总值． 最小值$$100\\div 10=10$$次，共减$$10\\times 27$$； 最多$$14$$次，共减去$$14\\times 27$$， 因此差为$$(14-10)\\times 27=108$$． " ]

[ "2017年第13届湖北武汉新希望杯六年级竞赛决赛第2题" ]

[ [ { "aoVal": "A", "content": "$$148$$~ " } ], [ { "aoVal": "B", "content": "$$149$$~ " } ], [ { "aoVal": "C", "content": "$$150$$~ " } ], [ { "aoVal": "D", "content": "$$151$$ " } ] ]

[ "拓展思维->拓展思维->行程模块->直线型行程问题->路程速度时间->平均速度->公式法" ]

[ "$$11:16-22:54$$共$$11$$小时$$38$$分，其中停车$$56$$分钟，运行$$10$$小时$$42$$分钟．$$1593\\div 10.7\\approx 149$$千米/时． " ]

[ "2016年IMAS小学高年级竞赛第一轮检测试题第7题3分" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$2$$ " } ], [ { "aoVal": "C", "content": "$$6$$ " } ], [ { "aoVal": "D", "content": "$$8$$ " } ], [ { "aoVal": "E", "content": "$$10$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->因数与倍数->因数个数定理->因数个数定理正应用->总个数" ]

[ "$$(192,120)={{2}^{3}}\\times 3$$，所以因数个数有$$(3+1)\\times (1+1)=8$$个． " ]

[ "2020年新希望杯二年级竞赛初赛（团战）第54题" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$4$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$6$$ " } ], [ { "aoVal": "E", "content": "$$9$$ " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "二年级一班有$$25$$人，其中男生有$$11$$人，则女生有$$25-11=14$$人，$$8$$人属兔，则属龙的有$$25-8=17$$人，女生中至少有$$14-8=6$$人属龙． " ]

[ "2013年全国迎春杯三年级竞赛初赛第3题" ]

[ [ { "aoVal": "A", "content": "$$84$$ " } ], [ { "aoVal": "B", "content": "$$85$$ " } ], [ { "aoVal": "C", "content": "$$86$$ " } ], [ { "aoVal": "D", "content": "$$87$$ " } ] ]

[ "知识标签->拓展思维->应用题模块->和差倍问题->和差问题->多量和差问题" ]

[ "设杰克分到的金币为$$1$$份量，则有 $$\\left. \\begin{matrix}1+11 1-15 1+20 \\end{matrix} \\right }\\Rightarrow 4+16=280\\Rightarrow 1=66\\Rightarrow 66+20=86$$ " ]

[ "2017年IMAS小学高年级竞赛（第二轮）第1题4分" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$18$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ], [ { "aoVal": "E", "content": "$$32$$ " } ] ]

[ "拓展思维->思想->对应思想", "Overseas Competition->知识点->应用题模块->周期问题" ]

[ "由图可知，从第一个三角形开始，以每五个三角形为一个周期，每个周期内有$$3$$个黑色三角形与$$2$$个白色三角形，即黑色三角形比白色三角形多$$1$$个．可知$$80$$个三角形共有$$16$$个周期， 所以黑色三角形比白色三角形总共多$$16$$个． 故选$$\\text{B}$$． " ]

[ "2007年第5届创新杯四年级竞赛第1题5分", "2007年四年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "225 " } ], [ { "aoVal": "B", "content": "223 " } ], [ { "aoVal": "C", "content": "221 " } ], [ { "aoVal": "D", "content": "219 " } ] ]

[ "拓展思维->拓展思维->计算模块->数列与数表->等差数列->等差数列求和" ]

[ "$$2007\\div 9-4\\text{=}219$$ " ]

[ "2016年第7届广东广州羊排赛六年级竞赛第7题1分" ]

[ [ { "aoVal": "A", "content": "$$96$$ " } ], [ { "aoVal": "B", "content": "$$97$$ " } ], [ { "aoVal": "C", "content": "$$98$$ " } ], [ { "aoVal": "D", "content": "$$99$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "数序至少要$$90\\times 3-88-85=97$$（分）． 故选$$\\text{B}$$． " ]

[ "2017年河南郑州豫才杯小学高年级六年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$1235810$$ " } ], [ { "aoVal": "B", "content": "$$1235811$$ " } ], [ { "aoVal": "C", "content": "$$1235812$$ " } ], [ { "aoVal": "D", "content": "$$1235813$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->数列与数表->数列规律->数列找规律填数" ]

[ "观察数列可发现：从第$$3$$项开始，每一项都等于前两项之和．所以第$$6$$项为$$5+8=13$$，即$$1235813$$． " ]

[ "2018年湖北武汉新希望杯五年级竞赛训练题（三）第5题" ]

[ [ { "aoVal": "A", "content": "$$116$$ " } ], [ { "aoVal": "B", "content": "$$120$$ " } ], [ { "aoVal": "C", "content": "$$136$$ " } ], [ { "aoVal": "D", "content": "$$150$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$$7\\times (1+3)=28$$（米），$$(420+280)\\times 2\\div 28=50$$（段），$$3\\times 50=150$$（棵）． " ]

[ "2014年全国迎春杯三年级竞赛初赛第1题" ]

[ [ { "aoVal": "A", "content": "$$5\\times99+1$$ " } ], [ { "aoVal": "B", "content": "$$100+25\\times4$$ " } ], [ { "aoVal": "C", "content": "$$88\\times4+37\\times4$$ " } ], [ { "aoVal": "D", "content": "$$100\\times0\\times5$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->整数->四则混合运算" ]

[ "A等于$$5\\times(100-1)+1=500-5+1=496$$，B等于$$100+100=200$$ ，C等于$$(88+37)\\times4=125\\times 4=500$$ ，D等于$$0$$． " ]

[ "2017年IMAS小学高年级竞赛（第一轮）第1题3分" ]

[ [ { "aoVal": "A", "content": "$$340$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{34}{2017}$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$20$$ " } ], [ { "aoVal": "E", "content": "$$34$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "$$\\frac{20\\times 17}{2+0+1+7}=\\frac{20\\times 17}{10}=2\\times 17=34$$． 故选$$\\text{E}$$． " ]

[ "2018年湖北武汉新希望杯六年级竞赛训练题（五）第2题" ]

[ [ { "aoVal": "A", "content": "$$3.16$$ " } ], [ { "aoVal": "B", "content": "$$3.15$$ " } ], [ { "aoVal": "C", "content": "$$3.14$$ " } ], [ { "aoVal": "D", "content": "$$3.13$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->方程基础->一元一次方程->分数、小数系数方程" ]

[ "$$\\frac{1}{4}\\pi {{d}^{2}}={{\\left( d-\\frac{1}{9}d \\right)}^{2}}$$，解得$$\\pi \\approx 3.16$$． " ]

[ "2014年IMAS小学高年级竞赛第一轮检测试题第1题3分" ]

[ [ { "aoVal": "A", "content": "$$8236$$ " } ], [ { "aoVal": "B", "content": "$$8506$$ " } ], [ { "aoVal": "C", "content": "$$8736$$ " } ], [ { "aoVal": "D", "content": "$$8836$$ " } ], [ { "aoVal": "E", "content": "$$9716$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "解法$$1$$：$$2015+1520+5201=8736$$，故选$$\\text{C}$$． 解法$$2$$：$$2015+1520+5201=2015+152+1520+5201-152=8888-152=8736$$，故选$$\\text{C}$$． " ]

[ "2021年鹏程杯六年级竞赛初赛第16题" ]

[ [ { "aoVal": "A", "content": "$$87$$ " } ], [ { "aoVal": "B", "content": "$$88$$ " } ], [ { "aoVal": "C", "content": "$$89$$ " } ], [ { "aoVal": "D", "content": "$$90$$ " } ], [ { "aoVal": "E", "content": "以上都不对 " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "设这五个老人的年龄为$$a$$，$$b$$，$$c$$，$$d$$，$$e$$，且$$a\\textless{}b\\textless{}c\\textless{}d\\textless{}e$$， 则$$e-a=6$$，$$a+b+c+d+e=85\\times 5=425$$． ①因为$$d\\leqslant e-1$$，$$c\\leqslant e-2$$，$$b\\leqslant e-3$$，$$a=e-6$$， 所以$$(e-6)+(e-3)+(e-2)+(e-1)+e\\geqslant 425$$， $$e\\geqslant 87.4$$； ②因为$$b\\geqslant a+1$$，$$c\\geqslant a+2$$，$$d\\geqslant a+3$$，$$e=a+6$$， $$a+(a+1)+(a+2)+(a+3)+(a+6)\\leqslant 425$$， $$a\\leqslant 82.6$$， 即$$e=a+6\\leqslant 88.6$$， 所以$$87.4\\leqslant e\\leqslant 88.6$$， 所以$$e=88$$． 故选$$\\text{B}$$． " ]

[ "2022年六年级竞赛（世界少年奥林匹克思维能力测评地方选拔活动）第4题" ]

[ [ { "aoVal": "A", "content": "$$\\frac58$$ " } ], [ { "aoVal": "B", "content": "$$\\frac56$$ " } ], [ { "aoVal": "C", "content": "$$\\frac14$$ " } ], [ { "aoVal": "D", "content": "$$\\frac38$$ " } ], [ { "aoVal": "E", "content": "$$\\frac23$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->分数" ]

[ "【分析】假设全班一共$$120$$人，按规定需$$\\frac34$$的选票才能当选，也就是至少需要得到$$90$$票，现在已经得到当选票数的$$\\frac56$$，求出已获得的选票数量，求出还需要的选票数量，再计算还需要的数量占余下票数的几分之几。【详解】假设全班一共$$60$$人；$$120\\times\\frac34=90$$（张）$$90\\times\\frac56=75$$（张）$$120\\times\\frac23=80$$（张）$$120-80=40$$（张）$$90-75=15$$（张）$$15\\div40=\\frac38$$【点睛】本题考查的是基础的分数乘除法应用题，找准单位``1''是解题的关键。 " ]

[ "2015年华杯赛四年级竞赛初赛", "2015年华杯赛三年级竞赛初赛", "2015年华杯赛四年级竞赛初赛", "2015年华杯赛三年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$\\text{3}$$ " } ], [ { "aoVal": "B", "content": "$$\\text{9}$$ " } ], [ { "aoVal": "C", "content": "$$\\text{11}$$ " } ], [ { "aoVal": "D", "content": "$$\\text{8}$$ " } ] ]

[ "拓展思维->拓展思维->计数模块->加乘原理->加法原理" ]

[ "设所用的$$\\text{1}$$元、$$\\text{2}$$元和$$\\text{5}$$元的人民币张数依次为$$x$$，$$y$$，$$z$$。由已知可得： $$x+2y+5z=18$$，$$x+y+z\\leqslant 10$$ 当$$z=0$$时，$$\\left( x\\text{，}y \\right)$$只能取$$\\left( 2\\text{，}8 \\right)$$，共$$\\text{1}$$种； 当$$z=1$$时，$$\\left( x\\text{，}y \\right)$$只能取$$\\left( 1\\text{，}6 \\right)$$，$$\\left( 3\\text{，}5 \\right)$$，$$\\left( 5\\text{，}4 \\right)$$，共$$\\text{3}$$种； 当$$z=2$$时，$$\\left( x\\text{，}y \\right)$$只能取$$\\left( 0\\text{，}4 \\right)$$，$$\\left( 2\\text{，}3 \\right)$$，$$\\left( 4\\text{，}2 \\right)$$，$$\\left( 6\\text{，}1 \\right)$$，$$\\left( 8\\text{，}0 \\right)$$，共$$\\text{5}$$种； 当$$z=3$$时，$$\\left( x\\text{，}y \\right)$$只能取$$\\left( 1\\text{，}1 \\right)$$，$$\\left( 3\\text{，}0 \\right)$$，共$$\\text{2}$$种。 综上可得，一共有：$$\\text{1}+\\text{3}+\\text{5}+\\text{2}=\\text{11}$$（种）不同的付款方式。 选C。 " ]

[ "2019年美国数学大联盟杯五年级竞赛初赛第3题5分" ]

[ [ { "aoVal": "A", "content": "$$0$$ " } ], [ { "aoVal": "B", "content": "$$1$$ " } ], [ { "aoVal": "C", "content": "$$2$$ " } ], [ { "aoVal": "D", "content": "$$4$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "$$1$$的任意次方都为$$1$$，所以$$1$$的$$2019$$次方等于$$1$$，$$1$$的$$2020$$次方等于$$1$$，$$1+1=2$$． 故选$$\\text{C}$$． " ]

[ "走美杯三年级竞赛", "走美杯四年级竞赛" ]

[ [ { "aoVal": "A", "content": "惠民超市一天卖出$$3$$盒保温杯，每盒装有$$2$$个杯子，每个杯子$$50$$元，一共卖了多少元 " } ], [ { "aoVal": "B", "content": "乐乐在长$$50$$米的游泳池里游了三个来回，他游了多少米 " } ], [ { "aoVal": "C", "content": "2箱蜜蜂一年可以产$$50$$千克的蜂蜜，照这样计算，$$3$$箱蜜蜂一年可以产多少千克的蜂蜜 " } ], [ { "aoVal": "D", "content": "悠悠一家三口人，每人每天早上各吃一个$$50$$克的鸡蛋，两天吃的鸡蛋共多少克 " } ] ]

[ "拓展思维->拓展思维->应用题模块->年龄问题->年龄问题之差倍型" ]

[ "略 " ]

[ "2013年IMAS小学中年级竞赛第二轮检测试题第2题4分" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$5$$ " } ], [ { "aoVal": "E", "content": "$$6$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->数列与数表->等差数列->等差数列的概念" ]

[ "从$$1$$号到$$7$$号选手的编码是从$$1$$到$$7$$的自然数，找到中间数是$$4$$，那么他们的和是$$4\\times 7=28$$，除了小明之外其他选手的编号和是$$26$$，所以小明的编号是$$28-26=2$$，即$$2$$号。选择A " ]

[ "2016年新希望杯六年级竞赛训练题（二）第6题" ]

[ [ { "aoVal": "A", "content": "$$3$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "设围巾$$x$$条，手套$$y$$条． $$\\begin{cases}70x+60y\\leqslant 500 x\\geqslant 2y\\geqslant 3 \\end{cases}$$ $$\\begin{cases}x=2 y=3 \\end{cases}$$，$$\\begin{cases}x=2 y=4 \\end{cases}$$，$$\\begin{cases}x=2 y=5 \\end{cases}$$，$$\\begin{cases}x=2 y=6 \\end{cases}$$，$$\\begin{cases}x=3 y=3 \\end{cases}$$，$$\\begin{cases}x=3 y=4 \\end{cases}$$，$$\\begin{cases}x=4 y=3 \\end{cases}$$，$$7$$种． " ]

[ "2017年第16届湖北武汉创新杯五年级竞赛邀请赛训练题（二）" ]

[ [ { "aoVal": "A", "content": "$$92$$ " } ], [ { "aoVal": "B", "content": "$$93$$ " } ], [ { "aoVal": "C", "content": "$$94$$ " } ], [ { "aoVal": "D", "content": "$$95$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "$$90\\times 3=270$$（分），甲，乙，丙总分是$$270$$分， $$88\\times 2=176$$（分），甲，丙总分是$$176$$分， $$270-176=94$$（分），丙的得分是$$94$$分． 故选$$\\text{C}$$． " ]

[ "2008年第6届创新杯五年级竞赛初赛A卷第5题5分", "2008年五年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "2 " } ], [ { "aoVal": "B", "content": "3 " } ], [ { "aoVal": "C", "content": "4 " } ], [ { "aoVal": "D", "content": "5 " } ] ]

[ "拓展思维->拓展思维->应用题模块->周期问题->基本排列的周期问题" ]

[ "设$$1$$，$$2$$，$$3$$，$$4$$，$$5$$，$$4$$，$$3$$，$$2$$为一节，每节$$8$$个数，由于$$2000=8\\times 250$$，所以，第2000个学生所报的数是第$$250$$节的最后一个数，因此，第2000个学生所报的数为$$2$$. " ]

[ "走美杯三年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$21$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$10$$ " } ], [ { "aoVal": "D", "content": "$$13$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->年龄问题->年龄问题之差倍型" ]

[ "再过$$4$$年毛毛的年龄将是泡泡年龄的$$2$$倍，所以那时泡泡的年龄为$$7$$岁，毛毛为$$7\\times 2=14$$(岁)，他们今年的年龄总和为$$7+14-4\\times 2=13$$(岁)。 " ]

[ "2008年第6届创新杯六年级竞赛初赛A卷第7题5分", "2008年六年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "80-\\/-90 " } ], [ { "aoVal": "B", "content": "68-\\/-78 " } ], [ { "aoVal": "C", "content": "56-\\/-62 " } ], [ { "aoVal": "D", "content": "95-\\/-102 " } ] ]

[ "拓展思维->拓展思维->应用题模块->列方程解应用题" ]

[ "设现有人数为$$x$$，则糖块数为$$5x+10$$，依题意得$$5x+10=4\\times1.5x-2$$，解得$$x=12$$，所以糖块数为$$5\\times 12+10=70$$.即在这些糖在68-\\/-78块之间. " ]

[ "2021年鹏程杯六年级竞赛初赛第7题" ]

[ [ { "aoVal": "A", "content": "$$234$$ " } ], [ { "aoVal": "B", "content": "$$286$$ " } ], [ { "aoVal": "C", "content": "$$534$$ " } ], [ { "aoVal": "D", "content": "$$654$$ " } ], [ { "aoVal": "E", "content": "以上都不对 " } ] ]

[ "拓展思维->能力->逻辑分析->代数逻辑推理" ]

[ "先从个位数考虑，有$$2\\times 2=4$$，$$2\\times 3=6$$，$$2\\times 6=12$$，$$2\\times 7=14$$．再考虑乘数的百位只能是$$2$$或$$3$$，因此只有三种可能的填法： $$2\\times 273=546$$， $$2\\times 327=654$$， $$2\\times 267=534$$， 其中只有$$546$$能被$$13$$整除，因此这个积是$$546$$． 故选$$\\text{E}$$． " ]

[ "2012年全国美国数学大联盟杯小学高年级竞赛初赛第32题", "2013年美国数学大联盟杯小学高年级竞赛初赛第32题5分" ]

[ [ { "aoVal": "A", "content": "$$2.75$$ " } ], [ { "aoVal": "B", "content": "$$3.25$$ " } ], [ { "aoVal": "C", "content": "$$3.75$$ " } ], [ { "aoVal": "D", "content": "$$4.5$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->平均数问题->公式类->直接求平均数" ]

[ "$$(1.75+7.25)\\div2=4.5$$. " ]

[ "2008年六年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$2$$场 " } ], [ { "aoVal": "B", "content": "$$3$$场 " } ], [ { "aoVal": "C", "content": "$$4$$场 " } ], [ { "aoVal": "D", "content": "$$5$$场 " } ] ]

[ "拓展思维->拓展思维->计算模块->方程基础->一元一次方程->分数、小数系数方程" ]

[ "解法一：张扬已经赢了$$20\\times 95 \\%=19$$（场）比赛，只输了$$1$$场，为了赢得$$96 \\%$$的比赛，他输的场数只能占总场数的$$4 \\%$$，$$1\\div 4 \\%=25$$（场），所以他至少还要赢$$25-20=5$$（场）。 解法二：在$$20$$场比赛中，张扬已经赢了$$20\\times 95 \\%=19$$（场），如果他以后每一场都赢，设他再赢$$x$$场，则$$19+x=\\left( 20+x \\right)\\times 96 \\%$$，解得$$x=5$$。 " ]

[ "2008年六年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "$$\\frac{1}{7}$$ " } ], [ { "aoVal": "B", "content": "$$\\frac{2}{15}$$ " } ], [ { "aoVal": "C", "content": "$$\\frac{1}{8}$$ " } ], [ { "aoVal": "D", "content": "$$\\frac{15}{112}$$ " } ] ]

[ "拓展思维->拓展思维->计算模块->比较与估算->比较大小综合->分数通分法->通分子" ]

[ "由于$$\\frac{1}{7}=\\frac{2}{14}\\textgreater\\frac{2}{15}\\textgreater\\frac{2}{16}=\\frac{1}{8}$$，$$\\frac{15}{112}\\textgreater\\frac{14}{112}=\\frac{1}{8}$$，所以$$\\frac{1}{8} \\textless{} \\frac{1}{7}$$，$$\\frac{1}{8} \\textless{} \\frac{2}{15}$$，$$\\frac{1}{8} \\textless{} \\frac{15}{112}$$因此，这些分数中最小的是$$\\frac{1}{8}$$. " ]

[ "2014年全国迎春杯五年级竞赛复赛第11题" ]

[ [ { "aoVal": "A", "content": "$$8$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$4$$ " } ], [ { "aoVal": "D", "content": "$$2$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "考虑到一定会有进位，退位．设原数数字和为$$a$$，则$$-3$$，$$+4$$定不是差$$7$$，否则无法成为连续$$4$$个自然数．$$\\div 5$$说明末位为$$0$$或$$5$$，当末位为$$5$$时，$$-3$$，$$+4$$均不进位退位．当末位为$$0$$时，$$-3$$退位，符合． 所以$$-3$$相当于数字和多$$6$$，$$a+6$$；$$+4$$相当于数字和多$$4$$，$$a+4$$；$$\\div 5$$ 相当于数字和$$\\times 2$$，$$a\\times 2$$；$$a\\times 2$$、$$a+2$$、$$a+4$$连续，$$a\\times 2$$为$$a+7$$，$$a+5$$，$$a+3$$中的一个． 分类讨论得到$$a\\times 2=a+5$$成立，所以$$a=5$$，数字和为$$5$$，尾数为$$0$$的有，$$500$$（舍弃），$$410$$，$$320$$，$$230$$，$$140$$，共$$4$$个． " ]

[ "2019~2020学年陕西宝鸡渭滨区五年级上学期期末第22题1分", "2019~2020学年山东济南高新区五年级下学期期末第10题1分", "2020年广东深圳龙岗区亚迪学校迎春杯五年级竞赛模拟第20题2分" ]

[ [ { "aoVal": "A", "content": "$$21$$，$$31$$ " } ], [ { "aoVal": "B", "content": "$$13$$，$$33$$ " } ], [ { "aoVal": "C", "content": "$$35$$，$$45$$ " } ], [ { "aoVal": "D", "content": "$$37$$，$$97$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "$$\\text{A}$$中$$21$$不是质数，故$$\\text{A}$$错误； $$\\text{B}$$中$$33$$不是质数，故$$\\text{B}$$错误． $$\\text{C}$$中$$35$$，$$45$$都不是质数，故$$\\text{C}$$错误． 故选$$\\text{D}$$． " ]

[ "2018年湖北武汉新希望杯五年级竞赛训练题（一）第4题" ]

[ [ { "aoVal": "A", "content": "$$48.2$$千克 " } ], [ { "aoVal": "B", "content": "$$49$$千克 " } ], [ { "aoVal": "C", "content": "$$50$$千克 " } ], [ { "aoVal": "D", "content": "$$52$$千克 " } ] ]

[ "拓展思维->拓展思维->应用题模块->平均数问题->公式类->加权平均数" ]

[ "$$48+(48.1-48)\\times 40=52$$（千克）． " ]

[ "2008年第6届创新杯六年级竞赛复赛第2题4分", "2008年六年级竞赛创新杯" ]

[ [ { "aoVal": "A", "content": "2006 " } ], [ { "aoVal": "B", "content": "2007 " } ], [ { "aoVal": "C", "content": "2008 " } ], [ { "aoVal": "D", "content": "2009 " } ] ]

[ "拓展思维->拓展思维->计算模块->定义新运算->直接运算型->普通型" ]

[ "由$$\\text{min}\\left( 2006\\text{，}2008 \\right)=2006$$，$$\\min \\left( 2007\\text{，}2009 \\right)=2007$$得，原式$$=\\max \\left( 2006\\text{，}2007 \\right)=2007$$ " ]

[ "2018年美国数学大联盟杯五年级竞赛初赛第33题5分" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$${{2}^{49}}$$ " } ], [ { "aoVal": "C", "content": "$${{2}^{50}}$$ " } ], [ { "aoVal": "D", "content": "$${{2}^{100}}$$ " } ] ]

[ "拓展思维->思想->转化与化归的思想" ]

[ "$$\\textasciitilde\\textasciitilde\\textasciitilde\\textasciitilde\\left(2^{2} \\times 2^{4} \\times 2^{6}\\times \\ldots ~\\times 2^{98} \\times 2^{100}\\right) \\div \\left(2^{1} \\times 2^{3} \\times 2^{5}\\times \\ldots ~\\times 2^{97} \\times 2^{99}\\right)$$ $$=\\left( {{2}^{2}}\\div {{2}^{1}} \\right)\\times \\left( {{2}^{4}}\\div {{2}^{3}} \\right)\\times \\left( {{2}^{6}}\\div {{2}^{5}} \\right)\\times \\ldots \\times \\left( {{2}^{98}}\\div {{2}^{97}} \\right)\\times \\left( {{2}^{100}}\\div {{2}^{99}} \\right)$$ $$=\\underbrace{2\\times 2\\times 2\\times \\cdots \\times 2}_{50个2}$$ $$={{2}^{50}}$$． 故选$$\\text{C}$$． " ]

[ "2018年第22届广东世界少年奥林匹克数学竞赛二年级竞赛决赛第2题5分" ]

[ [ { "aoVal": "A", "content": "$$41$$ " } ], [ { "aoVal": "B", "content": "$$51$$ " } ], [ { "aoVal": "C", "content": "$$61$$ " } ] ]

[ "拓展思维->能力->实践应用" ]

[ "根据重叠问题，可知这个班上共有同学：$$20+31-10=41$$人． 故选择$$\\text{A}$$． " ]

[ "2018年华杯赛小学中年级竞赛初赛第4题10分", "2018年第23届华杯赛小学中年级竞赛初赛第4题" ]

[ [ { "aoVal": "A", "content": "$$18$$ " } ], [ { "aoVal": "B", "content": "$$14$$ " } ], [ { "aoVal": "C", "content": "$$12$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "可以考虑纵向每列白色棋子数相等，故白棋子数量一定是$$6$$的倍数；考虑横向，每行棋子数不同，且最少为$$0$$，最多为$$6$$，白色棋子数最少为$$0+1+2+3+4+5=15$$，最多是$$1+2+3+4+5+6=21$$，故$$15\\sim 21$$间能被$$6$$整除的数只有$$18$$，故白棋子数量为$$18$$，黑棋子有$$36-18=18$$． ", "<p>首先考虑纵向，每列白色棋子数相等，故白棋子数量一定是$$6$$的倍数，</p>\n<p>考虑横向每行棋子数不同，且最少为$$0$$，最多为$$6$$，</p>\n<p>白棋子数量最少为$$0+1+2+3+4+5=15$$，</p>\n<p>最多为$$1+2+3+4+5+6=21$$，</p>\n<p>易知$$15-21$$间能被$$6$$整除的只有$$18$$，</p>\n<p>故白棋子数量为$$18$$，</p>\n<p>黑棋子数量为$$6\\times 6-18=18$$．</p>\n<p>故选$$\\text{A}$$．</p>" ]

[ "2015年湖北武汉世奥赛六年级竞赛模拟训练题（二）第1题" ]

[ [ { "aoVal": "A", "content": "$$76$$ " } ], [ { "aoVal": "B", "content": "$$77$$ " } ], [ { "aoVal": "C", "content": "$$78$$ " } ], [ { "aoVal": "D", "content": "$$79$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "这个十三位数只能是$$9979989991000$$． " ]

[ "2012年第10届创新杯四年级竞赛初赛第6题6分" ]

[ [ { "aoVal": "A", "content": "$$11$$ " } ], [ { "aoVal": "B", "content": "$$16$$ " } ], [ { "aoVal": "C", "content": "$$21$$ " } ], [ { "aoVal": "D", "content": "$$26$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "假设玻璃镜框全部完好，玻璃厂应付$$1000\\times 5=5000$$元，与实际情况相比少付了$$5000-4475=525$$元，每打碎一个镜框少付$$5+20=25$$元，因此一共打碎了$$525\\div 25=21$$个． 故选：$$\\text{C}$$． " ]

[ "2005年第3届创新杯六年级竞赛初赛第2题" ]

[ [ { "aoVal": "A", "content": "$$7$$ " } ], [ { "aoVal": "B", "content": "$$7\\frac{1}{2}$$ " } ], [ { "aoVal": "C", "content": "$$7\\frac{1}{3}$$ " } ], [ { "aoVal": "D", "content": "$$7\\frac{1}{4}$$ " } ] ]

[ "拓展思维->拓展思维->应用题模块->工程问题->合作工程问题->接力施工问题" ]

[ "甲乙合作完成需要： $$1\\div \\left( \\frac{1}{6}+\\frac{1}{10} \\right)$$ $$=1\\div \\frac{4}{15}$$ $$=3.75$$（小时）， 每人工作$$3$$小时，还剩下： $$1-\\left( \\frac{1}{6}+\\frac{1}{10} \\right)\\times 3$$ $$=1-\\frac{4}{5}$$ $$=\\frac{1}{5}$$， 甲再工作$$1$$小时，剩下的由乙完成需要： $$\\left( \\frac{1}{5}-\\frac{1}{6} \\right)\\div \\frac{1}{10}$$ $$=\\frac{1}{30}\\div \\frac{1}{10}$$ $$=\\frac{1}{3}$$（小时）， 一共$$3\\times 2+1+\\frac{1}{3}=7\\frac{1}{3}$$（小时）． 答：需要$$7\\frac{1}{3}$$时完成． 故选$$\\text{C}$$． " ]

[ "2017年全国希望杯小学高年级五年级竞赛初赛考前100题" ]

[ [ { "aoVal": "A", "content": "甲店 " } ], [ { "aoVal": "B", "content": "乙店 " } ], [ { "aoVal": "C", "content": "都一样 " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "设此商品的单价为$$a$$元，``买四赠一''即用$$4$$件的钱买了$$5$$件商品，购买单价为$$\\frac{4}{5}a=0.8a$$；``优惠$$\\frac{1}{4}$$''表示每件单价为$$\\left( 1-\\frac{1}{4} \\right)a=0.75a$$，故选择乙店． " ]

[ "2016年广东深圳四年级模拟考试鹏程杯集训", "2011年北京五年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$37800$$ " } ], [ { "aoVal": "B", "content": "$$38800$$ " } ], [ { "aoVal": "C", "content": "$$39900$$ " } ], [ { "aoVal": "D", "content": "太难了，容我在喝一杯$$82$$年的雪碧压压惊 " } ] ]

[ "知识标签->学习能力->七大能力->运算求解" ]

[ "甲、丙$$6$$分钟相遇的路程：$$\\left( {100 + 75} \\right) \\times 6 = 1050$$$$($$米$$)$$； 甲、乙相遇的时间为：$$1050 \\div \\left( {80 - 75} \\right) = 210$$$$($$分钟$$)$$； 东、西两村之间的距离为：$$\\left( {100 + 80} \\right) \\times 210 = 37800$$$$($$米$$)$$． " ]

[ "2018年第6届湖北长江杯五年级竞赛初赛A卷第6题3分" ]

[ [ { "aoVal": "A", "content": "$$8:00$$ " } ], [ { "aoVal": "B", "content": "$$18:40$$ " } ], [ { "aoVal": "C", "content": "$$19:20$$ " } ], [ { "aoVal": "D", "content": "$$20:00$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->因数与倍数->公因数与公倍数->因倍应用题->倍数应用题" ]

[ "$$10$$和$$8$$的最小公倍数是$$40$$， $$40\\times20=800$$（分）， $$6$$时$$20$$分再过$$800$$分钟为$$20$$时， 所以第$$20$$次同时发车为$$20$$时． 故选$$\\text{D}$$． " ]

[ "2017年全国美国数学大联盟杯小学高年级五年级竞赛初赛第29题" ]

[ [ { "aoVal": "A", "content": "$$48$$ " } ], [ { "aoVal": "B", "content": "$$49$$ " } ], [ { "aoVal": "C", "content": "$$60$$ " } ], [ { "aoVal": "D", "content": "$$61$$ " } ] ]

[ "知识标签->课内知识点->数学广角->鸽巢问题->利用抽屉原理解决实际问题" ]

[ "一个房间需要的最少多少人，才能使得总是有至少五个人是同一个月份出生的？ the minimum number：最小值；$$at$$ least：最少；$$born$$ in：出生；$$the$$ same month：同一个月份； $$4\\times 12+1=49$$，所以选$$\\text{B}$$． " ]

[ "2014年全国迎春杯五年级竞赛初赛第6题", "2014年全国迎春杯六年级竞赛初赛第6题", "2016年陕西西安小升初交大附中入学真卷9第11题", "2014年北京六年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$6$$ " } ], [ { "aoVal": "B", "content": "$$28$$ " } ], [ { "aoVal": "C", "content": "$$56$$ " } ], [ { "aoVal": "D", "content": "$$70$$ " } ] ]

[ "拓展思维->思想->方程思想" ]

[ "设丁拿了$$a$$件礼物，则四人花同样的钱，每人可以拿到$$a+\\frac{3+7+14}{4}=a+6$$件礼物， 实际情况：丁少拿了$$6$$件，乙多拿了$$1$$件，给丁$$14$$元，则货物单价$$14$$元， 丙多拿了$$14-6=8$$件，$$3$$件给甲，$$5$$件给丁，$$5\\times14=70$$元． " ]

[ "2019年第7届湖北长江杯六年级竞赛复赛B卷第1题3分" ]

[ [ { "aoVal": "A", "content": "$$16:05$$ " } ], [ { "aoVal": "B", "content": "$$17:55$$ " } ], [ { "aoVal": "C", "content": "$$18:00$$ " } ], [ { "aoVal": "D", "content": "$$18:05$$ " } ] ]

[ "拓展思维->能力->运算求解" ]

[ "本题中的相等关系是： 这只手表慢的时间$$-$$手表每小时比准确时间慢$$5$$分钟$$\\times $$标准时间经过的时间$$=0$$， 设标准时间经过了$$x$$小时，根据等量关系列方程求解即可． $$5+12=17$$时， 设标准时间经过了$$x$$小时，则 $$\\left( 6+x-17 \\right)\\times 60-5x=0$$， $$60\\left( x-11 \\right)-5x=0$$， $$60x-660-5x=0$$， $$55x=660$$， $$x=12$$． $$6:00+12=18:00$$． 所以准确时间应该是$$18:00$$． 故选$$\\text{C}$$． " ]

[ "2020年迎春杯六年级竞赛第10题12分" ]

[ [ { "aoVal": "A", "content": "$$23$$ " } ], [ { "aoVal": "B", "content": "$$29$$ " } ], [ { "aoVal": "C", "content": "$$41$$ " } ], [ { "aoVal": "D", "content": "$$61$$ " } ], [ { "aoVal": "E", "content": "$$97$$ " } ] ]

[ "拓展思维->能力->数感认知->数学概念理解（数）", "海外竞赛体系->知识点->组合模块->逻辑推理" ]

[ "所有的两位质数：$$11$$、$$31$$、$$41$$、$$61$$、$$71$$； $$13$$、$$23$$、$$43$$、$$53$$、$$73$$、$$83$$； $$17$$、$$37$$、$$47$$、$$67$$、$$97$$； $$19$$、$$29$$、$$59$$、$$79$$、$$89$$． 丙：丙说话之前甲只知道个位，就知道乙猜不出， 所以个位不可能是$$7$$， 丙手上拿的是数字和，通过数字和来确定这个信息的， 因此数字和不可能是$$8$$、$$10$$、$$11$$、$$13$$、$$16$$，将数字和为这些的都排除， 只剩下$$9$$个数：$$11$$、$$13$$、$$23$$、$$31$$、$$41$$、$$43$$、$$59$$、$$61$$、$$89$$． 乙：乙说话之前，丁拿的是数字差，甲一定猜不出， 所以排除$$11$$、$$41$$、$$59$$、$$61$$，那么个位不可能是$$1$$和$$9$$， 继续排除$$31$$、$$59$$、$$89$$． 剩下$$3$$个数：$$13$$、$$23$$、$$43$$， 乙通过十位确定甲一定知道丁猜不出，那么十位不可能是$$1$$和$$4$$， 故这个质数一定是$$23$$． " ]

[ "2011年全国世奥赛三年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "$$101$$ " } ], [ { "aoVal": "B", "content": "$$150$$ " } ], [ { "aoVal": "C", "content": "$$149$$ " } ], [ { "aoVal": "D", "content": "$$151$$ " } ] ]

[ "拓展思维->能力->运算求解->程序性计算" ]

[ "观察$$100$$和$$1$$之间的数，将$$\\left( 99 - 98\\right)$$、$$\\left( 97 - 96 \\right)$$、$$\\left( 95 - 94 \\right)$$、$$ \\cdots $$、$$\\left( 3 - 2 \\right)$$进行组合（ 除去$${1}$$和$${100}$$后，奇数前面是加号，偶数前面是减号），所以一共有$$\\left( 100 - 2 \\right) \\div 2 = 49$$个$$1$$，所以原式$$ = 100 + 1 + 49 = 150$$． " ]

[ "2015年华杯赛六年级竞赛初赛" ]

[ [ { "aoVal": "A", "content": "甲、乙 " } ], [ { "aoVal": "B", "content": "丙、丁 " } ], [ { "aoVal": "C", "content": "甲、丙 " } ], [ { "aoVal": "D", "content": "乙、丁 " } ] ]

[ "拓展思维->思想->逐步调整思想" ]

[ "由丙的话可知：丙没有捐款；再由甲的话可知：丁捐了款；再由乙的话可知：丁、甲之中最多有$$1$$人捐款．由此推知，甲没有捐款，乙捐了款，捐款的两人为乙和丁，选$$\\text{D}$$． " ]

[ "2013年全国迎春杯小学中年级竞赛复赛第2题" ]

[ [ { "aoVal": "A", "content": "一 " } ], [ { "aoVal": "B", "content": "二 " } ], [ { "aoVal": "C", "content": "三 " } ] ]

[ "知识标签->学习能力->七大能力->逻辑分析" ]

[ "第一个和第三个都说自己是疯子，所以他们不是牧师，只有第二个是牧师．而牧师说的是真话，所以第一个不是疯子．因此第三个才是真正的疯子． " ]

[ "2019年第22届世界少年奥林匹克数学竞赛四年级竞赛决赛（中国区）第1题2分" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$12$$ " } ], [ { "aoVal": "C", "content": "$$14$$ " } ], [ { "aoVal": "D", "content": "$$16$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "首先要求出工作总量，即$$15\\times 4\\times 18$$，再用工作总量除以工作人数，再除以每天的工作时间，即为所需要的天数．解决此题的关键是先求出总工作量． $$15\\times 4\\times 18\\div \\left[ \\left( 15+3 \\right)\\times \\left( 4+1 \\right) \\right]$$ $$=1080\\div 90$$ $$=12$$（天）． 答：这项工程$$12$$天完成． 故选 $$\\text{B}$$． " ]

[ "2017年全国美国数学大联盟杯小学高年级五年级竞赛初赛第33题" ]

[ [ { "aoVal": "A", "content": "$$1$$ " } ], [ { "aoVal": "B", "content": "$$3$$ " } ], [ { "aoVal": "C", "content": "$$5$$ " } ], [ { "aoVal": "D", "content": "$$7$$ " } ] ]

[ "拓展思维->思想->对应思想", "Overseas Competition->知识点->组合模块->逻辑推理->体育比赛" ]

[ "根据题意，$$F$$已经跟每个人都比赛过，包括$$G$$；同理，$$D$$，$$E$$也跟$$G$$比赛过，所以$$G$$比了$$3$$场 " ]

[ "2017年第16届湖北武汉创新杯五年级竞赛邀请赛训练题（四）" ]

[ [ { "aoVal": "A", "content": "周四 " } ], [ { "aoVal": "B", "content": "周五 " } ], [ { "aoVal": "C", "content": "周六 " } ], [ { "aoVal": "D", "content": "周日 " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "$$6$$月份有$$30$$天，按照每同$$7$$天计算，$$30\\div 7=4$$（周）$$\\cdots \\cdots 2$$（天），根据这个$$6$$月的周五比周四多$$1$$个，可知多余的$$2$$天必为周五和周六，即这个$$6$$月的最后一天是星期 六，所以$$7$$月$$1$$日是周日． " ]

[ "2017年第15届湖北武汉创新杯六年级竞赛决赛第6题" ]

[ [ { "aoVal": "A", "content": "$$5$$ " } ], [ { "aoVal": "B", "content": "$$6$$ " } ], [ { "aoVal": "C", "content": "$$8$$ " } ], [ { "aoVal": "D", "content": "$$10$$ " } ] ]

[ "拓展思维->拓展思维->数论模块->质数与合数->质数与合数判定->质数与合数的认识" ]

[ "数论．$$a$$、$$b$$互素，且计算结果分母为$$35$$，因此$$a$$、$$b$$为$$5$$和$$7$$，代入原式得$$\\frac{7m+5n}{35}=\\frac{31}{35}$$，则$$m=3$$，$$n=2$$，所以$$\\frac{5}{3}+\\frac{7}{2}-\\frac{1}{2\\times 3}=5$$． " ]

[ "2018年美国数学大联盟杯四年级竞赛初赛第32题5分" ]

[ [ { "aoVal": "A", "content": "$$2$$ " } ], [ { "aoVal": "B", "content": "$$5$$ " } ], [ { "aoVal": "C", "content": "$$7$$ " } ], [ { "aoVal": "D", "content": "$$9$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "$$9$$个连续正整数的和总是可以被整除． 设$$9$$个数中的中间数为$$a$$，则这$$9$$个数可表示为：$$a-4$$，$$a-3$$，$$a-2$$，$$a-1$$，$$a$$，$$a+1$$，$$a+2$$，$$a+3$$，$$a+4$$， 所以和为：$$\\left(a-4\\right)+\\left(a-3\\right)+\\left(a-2\\right)+\\left(a-1\\right)+a+\\left(a+1\\right)+\\left(a+2\\right)+\\left(a+3\\right)$$ $$+\\left(a+4\\right)=9a$$，即不管$$a$$为什么数时，其和必能被$$9$$整除． 故选：$$\\text{D}$$． " ]

[ "2016年全国中环杯四年级竞赛初赛第11题" ]

[ [ { "aoVal": "A", "content": "$$R\\textgreater T\\textgreater S\\textgreater Q\\textgreater P$$ " } ], [ { "aoVal": "B", "content": "$$R\\textgreater S\\textgreater T\\textgreater Q\\textgreater P$$ " } ], [ { "aoVal": "C", "content": "$$S\\textgreater R\\textgreater Q\\textgreater T\\textgreater P$$ " } ], [ { "aoVal": "D", "content": "$$S\\textgreater R\\textgreater T\\textgreater Q\\textgreater P$$ " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "$$Q+T=800$$①； $$Q+R=900$$②；$$P+T=700$$③；$$Q+S\\geqslant 1000$$④；$$R+T\\geqslant 1000$$⑤；由①②得：$$R\\textgreater T$$；由①③得：$$Q\\textgreater P$$；由②④得：$$S\\textgreater R$$；由②⑤得：$$T\\textgreater Q$$；所以：$$S\\textgreater R\\textgreater T\\textgreater Q\\textgreater P$$． " ]

[ "2017年河南郑州豫才杯五年级竞赛初赛第1题" ]

[ [ { "aoVal": "A", "content": "$$6$$万~~~~~~~~~~ " } ], [ { "aoVal": "B", "content": "$$0.06$$万~~~~ " } ], [ { "aoVal": "C", "content": "$$6000$$~~~~~~~ " } ], [ { "aoVal": "D", "content": "$$600$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "$$729636\\div 120\\approx 6000$$字． " ]

[ "2011年北京五年级竞赛" ]

[ [ { "aoVal": "A", "content": "$$0.216$$ " } ], [ { "aoVal": "B", "content": "$$0.432$$ " } ], [ { "aoVal": "C", "content": "$$0.144$$ " } ], [ { "aoVal": "D", "content": "$$0.288$$ " } ] ]

[ "拓展思维->思想->对应思想" ]

[ "第一箭射空，其他两箭射中的概率为$$\\left( 1-0.4 \\right)\\times 0.4\\times 0.4=0.096$$． 第二箭射空，其他两箭射中的概率为$$\\left( 1-0.4 \\right)\\times 0.4\\times 0.4=0.096$$． 第三箭射空，其他两箭射中的概率为$$\\left( 1-0.4 \\right)\\times 0.4\\times 0.4=0.096$$． 有两箭射空的概率为$$0.96+0.96+0.96=0.288$$． " ]

[ "2012年全国迎春杯六年级竞赛初赛第3题" ]

[ [ { "aoVal": "A", "content": "$$800$$ " } ], [ { "aoVal": "B", "content": "$$1000$$ " } ], [ { "aoVal": "C", "content": "$$1200$$ " } ], [ { "aoVal": "D", "content": "$$1400$$ " } ] ]

[ "知识标签->拓展思维->应用题模块->经济问题->基本经济概念->折扣问题" ]

[ "设进价为``$$1$$''，则第一天定价为``$$2$$''，第二天定价为``$$1.8$$''，最终售价为``$$1.44$$''． ``$$1.8$$''与``$$1.44$$''的差价等于$$360$$元，可知进价``$$1$$''$$=1000$$元． 设进价为$$x$$元． $$2x\\times (1-10 \\%)-360=1.44x$$， 解得：$$x=1000$$， 故答案为：$$1000$$． " ]

[ "2013年IMAS小学中年级竞赛第一轮检测试题第15题4分" ]

[ [ { "aoVal": "A", "content": "$$10$$ " } ], [ { "aoVal": "B", "content": "$$11$$ " } ], [ { "aoVal": "C", "content": "$$15$$ " } ], [ { "aoVal": "D", "content": "$$16$$ " } ], [ { "aoVal": "E", "content": "$$20$$ " } ] ]

[ "拓展思维->能力->构造模型->模型思想" ]

[ "可发现$$13+11+16=40$$恰为三只小兔子总共拔的萝卜数之$$2$$倍，因此三只小兔子共拔了$$40\\div 2=20$$根萝卜．故选$$\\text{E}$$． " ]

[ "2020年广东广州羊排赛三年级竞赛第14题8分" ]

[ [ { "aoVal": "A", "content": "一 " } ], [ { "aoVal": "B", "content": "四 " } ], [ { "aoVal": "C", "content": "五 " } ], [ { "aoVal": "D", "content": "六 " } ], [ { "aoVal": "E", "content": "日 " } ] ]

[ "拓展思维->拓展思维->应用题模块->周期问题->时间中的周期问题->日期中的周期" ]

[ "$$87$$集的电视连续剧在某个星期日开播，每周日、周一、周四、周五、周六都要播一集，五天一周期，$$87\\div 5=17$$（周）$$\\cdots \\cdots 2$$（天），所以最后一集在周一播出，选项$$\\text{A}$$正确． " ]

[ "2012年第8届全国新希望杯五年级竞赛复赛第4题" ]

[ [ { "aoVal": "A", "content": "$$3$$种 " } ], [ { "aoVal": "B", "content": "$$4$$种 " } ], [ { "aoVal": "C", "content": "$$5$$种 " } ], [ { "aoVal": "D", "content": "$$6$$种 " } ] ]

[ "拓展思维->能力->逻辑分析" ]

[ "$$\\overline{20\\square 12\\square }$$末位肯定为$$0$$，那么$$\\overline{20\\square 120}$$是$$3$$的倍数，中间的$$\\square $$有三种选择$$1、$$4$$、7$$，因此最后只有$$3$$种填法． " ]